During electrolysis, the mass of a substance liberated by the passage of an electric current is strictly determined by Faraday's Laws of Electrochemical Deposition. These laws state that:
1. "The amount of chemical change occasioned by the passage of an electric current is proportional to the quantity of electricity passed"; and PA1 2. "The masses of different substances liberated by a given quantity of electricity are proportional to their chemical equivalent weights."
The chemical equivalent weight of any substance is easily determined and remains a fixed standard for that substance under all conditions of electrolytic action. It is usually quoted in m.g.C.sup.-1, 1 Coulomb (C) being the quantity of electricity used when a current of one ampere is passed for one second.
If the chemical equivalent weight is represented by z, the mass, m, of any substance liberated during an electrolytic process is given by: EQU m=z.I.t (1)
where I is the current passed in amperes and t is the time in seconds.
During normal electrolytic processes, it is not possible to induce a current to flow through the electrolyte unless the voltage across the electrodes of the electrolytic cell is raised to some specific value, which varies according to the electrolyte and the electrode composition. This voltage, V.sub.d, is known as the Decomposition Voltage. Hitherto, it has not been possible to arrange for electrolytic cells to function at voltages sufficiently low to enable of very low-power inputs to the cell.
Any process which can be arranged to run in such a way that, when the calorific value of a liberated gas is higher than the power required to run the electrolytic process which liberates that gas, will act as a net provider of energy. The apparent surplus of energy coming, in this instance, from the bond dissociation energies of the ions involved in the process.
An example of the operation of an electrolytic cell will serve to illustrate the above points more clearly.
Let us first consider a cell which liberates hydrogen gas by the electrolysis of water containing a standard electrolyte such as H.sub.2 SO.sub.4 or Li.sub.2 SO.sub.4. If such a cell is run such that its terminal voltage is 5 volts and the current being passed through it is 2 amperes, it will require a power source of at least 10 watts, allowing for small losses in wires and contact resistances. The mass, and hence calorific value, of the hydrogen liberated from such a cell will be in accordance with Faraday's Laws and will be proportional to the product of current and time as outlined above. However, the product of current and time is not the same thing as the product of current and voltage, which gave us the power consumption of the cell. In the case of this cell, the power input is given simply by:
P.sub.in =V.times.I
where V is the cell voltage and I is the cell current.
To calculate the power output of such a cell, we need to know how much energy is available from a given mass of hydrogen gas when it combines with oxygen during combustion. This figure is 285 KJ.mol.sup.-1, where 1 KJ (kilojoule) is the energy converted when 1 kilowatt of power is used for a duration of 1 second. Since the chemical equivalent weight of hydrogen is known to be 0.01045 mgC.sup.-1, it can be calculated, according to (1) above, that the cell will yield a mass, m, of hydrogen gas given by EQU m=0.01045.times.10.sup.-3.times.2 g.s.sup.-1 EQU =2.09.times.10.sup.-5 g.s.sup.-1
1 mol of hydrogen gas, as molecular hydrogen H.sub.2, has a mass of 2.016 g. Utilising the energy content of hydrogen as it undergoes combustion, we therefore have an energy yield from the cell of: ##EQU1##
It can be seen, therefore, that this conventional cell only produces just over a quarter as much energy from the full combustion of its hydrogen yield as the electrical energy required to make it run. Such a device is not an efficient converter of energy.
Consider now the performance of the same cell if its current of 2 amperes were to flow using a very much smaller potential of only 0.5 volts. The input power is given by the same equation (2) above, namely: EQU P.sub.in =V.times.I
=0.5.times.2=1 W
The output power, however, remains the same as in the 5 volt example, it being dependent solely upon the parameters of current and time.
The 0.5 volt cell, therefore, yields a supply of hydrogen gas which is capable of being burned to provide some 2.9 times the electrical energy input to the cell.
In the past it has not been possible to cause electrolysis cells to operate at the small voltages necessary to achieve this kind of "energy multiplier" effect. The natural barrier of the established decomposition voltage always halted the process some way before the over-unity effects of the cell became evident.